Directions

The numbers next to each row and column indicate the size of groups of contiguous black pixels. Fill in the pixels.

For example: "2-1" in a five-square row could indicate any of three possible arrangements:

We're given a 21x21 grid of squares. The content of each row is described on the right side of the grid, and the content of each column is described at the bottom. Each number in the row/column description describes exactly that many black squares, while each "-" mark describes one or more white squares.

The number of total squares in any one row or column is exactly 21, so the sum of the count of black squares for any given row or column plus the sum of the "-" marks for that row or column must not be greater than 21. Furthermore, any row or column description whose sum of the count of black squares and "-" marks equals exactly 21 has been described to us in the final state. So let's look for any rows or columns that add up to 21 and go ahead and fill them.

As shown above, ROWS 3 and 7 and COLUMNS E and G were described in their exact state so all we had to do was fill them in as described. So now that we've exhausted our easy rows, let's move on to the next easiest case. Look for any rows or columns that add to 20, and draw the possible outcomes. If any squares overlap in all possible states for a given row or column, then we know that square must be black. Squares that exist in some but not all possibilities may or may not be black squares. We'll mark them with blue to keep them in mind.

ROW 1's description adds to 20. All possible solutions to ROW 1 are presented above. We'll mark any square that exists in all possible solutions in black, and we'll keep the remaining possibilities in blue. Now we'll transfer the result back to the overall puzzle and check our progress.

ROW 19 also adds to 20. We'll perform the same possibility check on it.

Now transfer it back to the overall puzzle.

That takes care of all the rows that add to 20. Now let's search for columns.

COLUMN D fits the bill. Once again, we perform possibility analysis and transfer the result to the overall puzzle.

We've now exhausted the rows and columns that add to 20. Let's keep going down this route and look for rows or columns that add to 19.

ROWS 4 and 5 are identical, so we'll take the solution from ROW 4 and apply it to ROWS 4 and 5 in the overall puzzle.

ROW 15 also adds to 19. Let's analyze possible solutions.

And again, add the result back into the overall puzzle.

We've now reached the point of tedium. Progressing along the brute-force path of mapping out possible outcomes now requires too much effort. It's time to find an alternate way to attack the puzzle.

Instead of examining an entire row or column at a time, let's look at portions of a row or column now that we've filled out some of the puzzle. Take COLUMN A, for example. We know the first contiguous group of black squares consists of 7 squares. We also know that squares A3, A4, A5, and A7 are black. Based on these two pieces of information, we can conclude that the first contiguous square of COLUMN A starts at A1 and ends at A7, starts at A3 and ends at A9, or something in between. Regardless, square A6 must be black.

Moving along, let's examine COLUMN D. We've already satisfied the 1-3-1 conditions at beginning and end, so all we need is a group of 4 black squares in the middle. Since D15 is a single square, D14 cannot be black and therefore must be white.

Let's suppose that square I1 is black. If so, that means that square I3 is the start of a group of 4 black squares. However, if we fill in squares I4, I5, and I6, we'll end up with a group of 5 black squares from I3 to I7. This violates the description of COLUMN I, so we have proven that I1 cannot be black and therefore must be white. I3 must be the first single black square in ROW I, and I7 is part of a 4-square group. Since I3 is a single black square, I4 cannot be black and therefore must be white.

COLUMN L starts with 4 black squares. Since L1 is black, the next three squares (L2, L3, and L4) must also be black. L5 cannot be black and therefore must be white.

COLUMN O starts with 7 black squares. Since O1 is black, the next six squares (O2, O3, O4, O5, O6, and O7) must also be black.

COLUMNS R and S start with a single black square, and the next group consists of 3 black squares. Since R3 and S3 are black, the subsequent two squares (R4, R5 and S4, S5, respectively) must be black.

COLUMN U contains an initial group of 7 black squares. We can analyze it the same as we analyzed COLUMN A, and we conclude that U4, U5, and U6 must all be black. These changes are depicted in the next image.

There isn't enough room between L1 and O1 for a group of 2 black squares plus white padding, so L1 must be part of a group of 2 black squares. Since I1 is white, J1 must be a single black square. Therefore K1 must be white and M1 finishes out the group of 2 with L1. N1 must then be white, and U1 must then in turn be black.

In ROW 4, L4 is a single black square, so K4 and M4 must be white. J4 must be the final single black square in ROW 4. Because the 1-3-1 at the end of ROWS 4 and 5 is satisfied, P4, T4 and P5, T5 must be white.

To finish out the initial group of 4 black squares in COLUMN J, J2 must be black, and J5 must be white.

U2 must be black to finish out the initial group of 7 black squares in COLUMN U.

By now, the pattern here should look pretty familiar. It's quite clearly a QR code. Maybe we can get some help if we do an internet search for 21x21 QR codes!

I found some helpful information at QRStuff.com. Specifically, I found the image below that describes QR code data standards.

Based on the QR code depiction, we can go ahead and fill in our top left, top right, and bottom left corners.

K2 must be black to complete ROW 2.

COLUMN C has a group of 2 black squares somewhere between C9 and C13. C12 cannot be black because the first group in ROW 12 contains only 22 black squares, and that condition has already been satisfied by D12 and E12. C13 cannot be black, because C12 above and C14 below must be white. C11 cannot be black, because the group of 5 black squares in ROW 11 cannot start that close to the left edge, because there isn't sufficient room for previous black squares and white space. Therefore C9 and C10 must be black to complete COLUMN C.

COLUMN F also has a group of 2 black squares somewhere between F9 and F13. F11, F12, and F13 must be white to satisfty conditions on ROWS 11, 12, and 13, respectively. Therefore F9 and F10 must be black to complete COLUMN F.

A9 and B9 must be black to satisfy the initial group of 3 black squares in ROW 9. This completes COLUMN A.

Now that we've satisfied the initial group of 3 black square in ROW 9, we know that E9 through G9 are part of a group of 8 black squares. H9 through L9 must also be black to complete that group of 8.

Because COLUMN G is completed, G10 cannot be used to satisfy the initial group of 5 black squares in ROW 10. Therefore B10 must be black. Now a single black square remains in COLUMN B, somewhere between B12 and B13. Again, as the first condition in ROW 12 has already been met, B12 must be white. Therefore B13 must be black to complete COLUMN B.

With the first two conditions of ROW 9 satisfied, D9 must be white. Therefore D13 must be black to complete COLUMN D.

So we've now finished the left third of puzzle and most of the top third. The rest is looking pretty sparse though. It looks like I missed another easy one: ROW 8 has two groups of 2 black squares. Based on the QR code depiction above, A8 through H8 and N8 through U8 must be white. That means that the two groups of 2 must exist between I8 and M8. That only works out if I8, J8 and L8, M8 are black, leaving K8 white in the middle to complete ROW 8.

Knowing that L8 and L9 are black, we can deduce that L10 must be black to satisfy the group of 3 black squares in COLUMN L.

We now know that I5 cannot be black without violating the description of COLUMN I, so I5 must be white. Therefore K5 and M5 must be black to complete ROW 5.

Because M5 is black, M6 must also be black to satisfy a group of 4 black squares in COLUMN M. This completes ROW 6. With ROW 6 completed, I10 must be black to satisfy a group of 4 black squares in COLUMN I.

Now that we've made some decent progress, let's go back and check out some possibilities. COLUMN K looks like a good place to start.

And transfer it back to the overall puzzle.

Now that we've satisfied the second group of 3 black squares in ROW 19, we know that I19 must be a single black square, J19 must be white, and N19 must be white.

Let's take a closer look at ROW 11. The last three groups consist of 2-5-2. There's not enough room for those three sets plus white squares to exist between M11 and U11, so some of the black squares must occur to the left of M11. I11 and L11 cannot be black without violating descriptions of COLUMNS I and L, respectively, so we can conclude that ROW 11 must contain a group of 2 black squares from J11 to K11. This satisfies a group of 4 black squares for COLUMN K, so K15 must be white.

Now let's look at COLUMN L. L11, L12, and L13 cannot be black and therefore must be white. That leaves exactly enough room to use L14 through L21 to complete COLUMN L.

ROW 21 requires that L21 be a single black square, so K21 must be white. Therefore K18 must be black to satisfy the group of 3 in COLUMN K. K17 must then be white and K16 must be black to complete COLUMN K.

The description of ROW 14 requires a single black square somewhere to the left of J14. Based on the QR code depiction above, A14 through H14 must be white. I14 must be the initial black square in ROW 14, and M14 must be black to satisfy the group of 3.

COLUMN J requires a group of 3 black squares at the end of the column. Since J14 and J19 cannot be black, that group of 3 has to exist somewhere between J15 and J18. We can conclude that J16 and J17 must be black, and the group is finished with either J15 or J18.

Now that we've determined J17 is black, we know that it must be part of a group of 2 black squares. I17 must also be black to finish this group.

Back to possibility analysis. Let's check out ROW 11.

And transfer it back to the overall puzzle.

Back to ROW 10. We need to place a group of 2 and a group of 3 black squares somewhere between N10 and U10. N10 and O10 cannot be black without violating the descriptions of COLUMNS N and O, respectively, so they must be white. That leaves exactly enough room to complete ROW 10 using the squares from P10 to U10.

I14 is part of a group of 4 black squares in COLUMN I. However, I14 cannot be the last in the group, since I11 must be white. Therefore I15 must also be black. This means that J15 must be white, and J18 must be black to complete COLUMN J.

M19 is part of a group of 3 black squares in COLUMN M. However, M19 cannot be the first in the group, since M21 must be white. Therefore M18 must also be black. This satisfies the group of 4 black squares in ROW 18, so I18 must be white. I16 and I20 through I21 must be black to complete COLUMN I.

Based on the QR depiction above, H16 must be white. Now that we know I16 is black, M16 must also be black and ROW 16 is complete.

Since M17 must be white to separate the groups of 4 and 3 black squares in COLUMN M, we know that M20 must be black to finish the group of 3, and M15 and M13 must be black to finish the group of 4. M10 cannot be black, so M11 must be black to complete COLUMN M.

With the group of 5 black squares finished in ROW 11, R11 must be white.

Since M15 is black, N15 must be white. That leaves exactly enough space from Q15 to U15 to complete ROW 15.

Q10 through Q11 and Q15 are all part of the same group of 6 black squares in COLUMN Q. We can fill in Q12 through Q14 to finish out that group.

P15 is part of a group of 3 black squares in COLUMN P. P16 must be white, so P13 and P14 must be black to finish the group of 3. There isn't enough room between P15 and P19 for a group of 2 black squares with white separators, so P19 must be part of a group of 2. In order to have room for the single black square at the end of COLUMN P, P20 must be white. P18 must be black to finish off the group of 2, and P21 is black to complete COLUMN P.

U10 is part of a group of 4 black squares in COLUMN U. However, U10 cannot be the last in the group, since U8 must be white. U11 and U12 must also be black. This completes ROW 11.

Since S11 must be white, S9 must be black to satisfy the group of 2 black squares in COLUMN S.

To complete ROW 14, one of the squares between S14 and T14 must be black. T14 cannot be black because T12 through T13 don't provide enough space for a group of 1 black square with white space on either side. Therefore S14 is black and this completes ROW 14.

S14 also completes COLUMN S, so S19 must be white. O19 must be black to satisfy the group of 4 black squares in ROW 19.

With O19 black, O18 must be white. Therefore Q18 must be black to satisfy the group of 2 black squares in ROW 18.

Q21 cannot be black without violating the description of COLUMN Q. O21 cannot be black because there is not enough room to squeeze another single black square to the left of it. Therefore P21 is a single black square and a group of 2 black squares must exist somewhere between R21 and U21. However, S21 cannot be black because COLUMN S is completed. T21 and U21 must be black to complete ROW 21.

With T21 black, T20 must also be black to satisfy the final group of 2 black squares in COLUMN T. This completes ROW 20. T19 must then be white, so U19 must be black to complete ROW 19.

With ROW 20 completed, Q17 must be black to complete COLUMN Q.

To complete ROW 18, a single black square must exist somewhere between S18 and T18. However, S18 can't be black because COLUMN S is completed. T18 must be black to complete ROW 18.

With T18 black, T17 must be black to complete COLUMN T.

With T17 black, Q17 must be part of a group of 2 black squares. Because COLUMN P is completed, R17 must be black to complete ROW 17. This also completes COLUMN R.

With COLUMNS R and T completed, S9 must be a single black square in ROW 9. To complete ROW 9, a group of 2 black squares must exist somewhere between N9 and Q9. The only two contiguous squares in uncompleted columns are N9 and O9, which must both be black to complete ROW 9. This also completes COLUMNS N and O.

With ROW 9 completed, U13 must be black to complete COLUMN U and ROW 13.

Finally, H12 must be black to complete ROW 12 and COLUMN H.

And here's the final solution, without the clutter.